Posted in Python onJune 23, 2018
本文实例讲述了Python实现重建二叉树的三种方法。分享给大家供大家参考,具体如下:
学习算法中,探寻重建二叉树的方法:
- 用input 前序遍历顺序输入字符重建
- 前序遍历顺序字符串递归解析重建
- 前序遍历顺序字符串堆栈解析重建
如果懒得去看后面的内容,可以直接点击此处本站下载完整实例代码。
思路
学习算法中,python 算法方面的资料相对较少,二叉树解析重建更少,只能摸着石头过河。
通过不同方式遍历二叉树,可以得出不同节点的排序。那么,在已知节点排序的前提下,通过某种遍历方式,可以将排序进行解析,从而构建二叉树。
应用上来将,可以用来解析多项式、可以解析网页、xml等。
本文采用前序遍历方式的排列,对已知字符串进行解析,并生成二叉树。新手,以解题为目的,暂未优化,未能体现 Python 简洁、优美。请大牛不吝指正。
首先采用 input 输入
节点类
class treeNode: def __init__(self, rootObj = None, leftChild = None, rightChild = None): self.key = rootObj self.leftChild = None self.rightChild = None
input 方法重建二叉树
def createTreeByInput(self, root): tmpKey = raw_input("please input a key, input '#' for Null") if tmpKey == '#': root = None else: root = treeNode(rootObj=tmpKey) root.leftChild = self.createTreeByInput(root.leftChild) root.rightChild = self.createTreeByInput(root.rightChild) return root
以下两种方法,使用预先编好的字符串,通过 list 方法转换为 list 传入进行解析
myTree 为实例化一个空树
调用递归方法重建二叉树
treeElementList = '124#8##5##369###7##' myTree = myTree.createTreeByListWithRecursion(list(treeElementList)) printBTree(myTree, 0)
递归方法重建二叉树
def createTreeByListWithRecursion(self, preOrderList): """ 根据前序列表重建二叉树 :param preOrder: 输入前序列表 :return: 二叉树 """ preOrder = preOrderList if preOrder is None or len(preOrder) <= 0: return None currentItem = preOrder.pop(0) # 模拟C语言指针移动 if currentItem is '#': root = None else: root = treeNode(currentItem) root.leftChild = self.createTreeByListWithRecursion(preOrder) root.rightChild = self.createTreeByListWithRecursion(preOrder) return root
调用堆栈方法重建二叉树
treeElementList = '124#8##5##369###7##' myTree = myTree.createTreeByListWithStack(list(treeElementList)) printBTree(myTree, 0)
使用堆栈重建二叉树
def createTreeByListWithStack(self, preOrderList): """ 根据前序列表重建二叉树 :param preOrder: 输入前序列表 :return: 二叉树 """ preOrder = preOrderList pStack = SStack() # check if preOrder is None or len(preOrder) <= 0 or preOrder[0] is '#': return None # get the root tmpItem = preOrder.pop(0) root = treeNode(tmpItem) # push root pStack.push(root) currentRoot = root while preOrder: # get another item tmpItem = preOrder.pop(0) # has child if tmpItem is not '#': # does not has left child, insert one if currentRoot.leftChild is None: currentRoot = self.insertLeft(currentRoot, tmpItem) pStack.push(currentRoot.leftChild) currentRoot = currentRoot.leftChild # otherwise insert right child elif currentRoot.rightChild is None: currentRoot = self.insertRight(currentRoot, tmpItem) pStack.push(currentRoot.rightChild) currentRoot = currentRoot.rightChild # one child is null else: # if has no left child if currentRoot.leftChild is None: currentRoot.leftChild = None # get another item fill right child tmpItem = preOrder.pop(0) # has right child if tmpItem is not '#': currentRoot = self.insertRight(currentRoot, tmpItem) pStack.push(currentRoot.rightChild) currentRoot = currentRoot.rightChild # right child is null else: currentRoot.rightChild = None # pop itself parent = pStack.pop() # pos parent if not pStack.is_empty(): parent = pStack.pop() # parent become current root currentRoot = parent # return from right child, so the parent has right child, go to parent's parent if currentRoot.rightChild is not None: if not pStack.is_empty(): parent = pStack.pop() currentRoot = parent # there is a leftchild ,fill right child with null and return to parent else: currentRoot.rightChild = None # pop itself parent = pStack.pop() if not pStack.is_empty(): parent = pStack.pop() currentRoot = parent return root
显示二叉树
def printBTree(bt, depth): ''''' 递归打印这棵二叉树,#号表示该节点为NULL ''' ch = bt.key if bt else '#' if depth > 0: print '%s%s%s' % ((depth - 1) * ' ', '--', ch) else: print ch if not bt: return printBTree(bt.leftChild, depth + 1) printBTree(bt.rightChild, depth + 1)
打印二叉树的代码,采用某仁兄代码,在此感谢。
input 输入及显示二叉树结果
解析字符串的结果
完整代码参见:https://github.com/flyhawksz/study-algorithms/blob/master/Class_BinaryTree2.py
希望本文所述对大家Python程序设计有所帮助。
Python实现重建二叉树的三种方法详解
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